Question

One factor of mc017-1. Jpg is (x â€" 2). What are all the roots of the function? Use the Remainder Theorem. X = 1, x = 2, or x = 4 x = â€"2, x = 1, or x = 2 x = 2, x = 4, or x = 16 x = â€"16, x = 2, or x = 16.

Answer 1

All the factors for f(x) is 4(x-2)(x+2)(x-1).

Given to us,

Function, $f(x)=4x^3-4x^2-16x+16$,

Factor 1, (x – 2)

According to remainder theorem, if any polynomial f(x) is divided by any factor of its own and remainder is zero then the quotient is the factor of that polynomial with a lower degree.

$\dfrac{f(x)}{x+a} = q(x)$

(x+a) and q(x) are factor of f(x) with lower degree.

Therefore,

$\dfrac{f(x)}{x-2}=\dfrac{4x^3-4x^2-16x+16}{x-2} = 4x^2+4x-8$

Further, $4x^2+4x-8$ can be written as,

$4x^2+4x-8\\=4(x^2+x-2)\\=4(x^2+2x-x-2)\\=4(x+2)(x-1)$

Hence, all the factors for f(x) is 4(x-2)(x+2)(x-1).

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